Now, let's understand why these statements hold true. Is it simply a matter of luck, or is there a deeper theory behind it?

The main observation that we need, is the way in which these values are related to each other along the lines:
When we go from right to left, we are multiplying by \( \tan \theta \). Conversely, when we go from left to right, we are dividing by \( \tan \theta \), which is equivalent to multiplying by \( \cot \theta \).
When we go from upper right to lower left, we are multiplying by \( \sin \theta \). Conversely, when we go from lower left to upper right, we are dividing by \( \sin \theta \), which is equivalent to multiplying by \( \csc \theta \).
When we go from upper left to lower right, we are multiplying by \( \cos \theta \). Conversely, when we go from lower right to upper left, we are dividing by \( \cos \theta \), which is equivalent to multiplying by \( \sec \theta \).

Reciprocal identities
The observation tells us that when we proceed in a fixed direction, we have a geometric progression. Hence, the product of the first and third term, is equal to the square of the second term. Thus, we get
\[ \tan \theta \times \cot \theta = 1^2 = 1 \Rightarrow \tan \theta = \frac{ 1 } { \cot \theta }. \]

Pythagorean identities
We start off with the fact that \( (\sin \theta)^2 + (\cos \theta)^2 = (1)^2 \), and record this in the \( 1 - \sin - \cos \) triangle.
How then do we obtain the other 2 identities? Consider what happens when we divide by each term by \( \cos \theta \).
\[ \left( \frac{ \sin \theta } { \cos \theta } \right) ^2 + \left( \frac{ \cos \theta } { \cos \theta } \right)^2 = \left( \frac{1}{ \cos \theta } \right)^2 \Leftrightarrow \tan ^2 \theta + 1 ^2 = \sec ^2 \theta. \]
In the grid, this is equivalent to moving each point from the lower right to the upper left. Thus, the \( 1 - \sin - \cos \) triangle will result in the \( \sec - \tan - 1 \) triangle.
Similarly, dividing each term by \( \sin \theta \) is equivalent to moving from the lower left to the upper right. Thus, the \( 1 - \sin - \cos \) triangle will shift to the \( \csc - 1 - \cot \) triangle.

Parity identities.
If we know that \( \sin \theta \) is odd and \( \cos \theta \) is even, and recall that when multiplying by an even function preserves its parity while multiplying by an odd function changes its parity, we get that functions along the upper left-lower right diagonal have the same parity, and this parity alternates as the line changes.